Question

Find the value of a, if the distance between the points P(3, -6) and Q(-3, a) is 10 units.

Pls Don’t Post Irrelevant answer :)

Answer 1

Given that , The points P(3, -6) and Q(-3, a) and the distance between the points is 10 units .

Exigency To Find : The Value of a ?

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⠀⠀▪︎⠀We know that to calculate the Distance between two points if we have given with co – ordinates of the two points ( x₁ , y₁ ) and ( x₂ , y₂ ) we use Distance Formula and that's given by :

$\qquad \star \:\:\underline {\boxed {\pmb{\sf { \:Distance \: =\: \: \sqrt{ \bigg( x_2 - x_1 \bigg)^2 + \bigg( y_2 - y_1 \bigg)^2 \:}}}}}$

⠀⠀⠀⠀⠀Here ,

• x₂ = – 3 ,

• x₁ = 3 ,

• y₂ = a ,

• y₁ = – 6 &,

• Distance between these points is 10 units .

$\\\\\\\qquad:\implies \sf \:Distance \: =\: \: \sqrt{ \bigg( x_2 - x_1 \bigg)^2 + \bigg( y_2 - y_1 \bigg)^2} \\ \\\\ \qquad:\implies \sf \:10 \: =\: \: \sqrt{ \bigg( (-3) - 3 \bigg)^2 + \bigg( ( a ) - ( – 6 ) \bigg)^2} \\\\\\ \qquad:\implies \sf \:10 \: =\: \: \sqrt{ \bigg( (-3) - 3 \bigg)^2 + \bigg( a + 6 \bigg)^2 }\\\\\\\qquad:\implies \sf \:10 \: =\: \: \sqrt{ \bigg( -6 \bigg)^2 + \bigg( a + 6 \bigg)^2} \\\\\\ \qquad:\implies \sf \:\bigg( 10 \bigg)^2 \: =\: \:\Bigg( \sqrt{ \bigg( -6 \bigg)^2 + \bigg( a + 6 \bigg)^2}\Bigg)^2 \\\\\\ \qquad:\implies \sf \:100\: =\: \:\ \bigg( -6 \bigg)^2 + \bigg( a + 6 \bigg)^2 \\\\\\ \qquad:\implies \sf \:100\: =\: \:\ \bigg( 36 \bigg) + \bigg( a + 6 \bigg)^2 \\\\\\ \qquad:\implies \sf \:100\: - 36 =\: \: \bigg( a + 6 \bigg)^2 \\\\\\ \qquad:\implies \sf \:64 =\: \: \bigg( a + 6 \bigg)^2 \\\\\\ \qquad:\implies \sf \sqrt{\:64 } =\: \: \sqrt{ \bigg( a + 6 \bigg)^2 } \\\\\\ \qquad:\implies \sf \:8 =\: \: \bigg( a + 6 \bigg) \\\\\\ \qquad:\implies \sf \:8 =\: \: a + 6 \\\\ \\ \qquad:\implies \sf \:a=\: \: 8 - 6 \\\\\\ \qquad:\implies \underline {\boxed {\pmb{\frak{ \:a =\: \: 2 }}}}\:\: \bigstar$

$\qquad \therefore \:\underline {\sf Hence, \:The \:Value \:of \:a \:is \:\pmb{\bf{2}}\:.}$